Re: problem mutex-thread "Unlocking mutex owned by another thread ???"

"James Kanze" <>
15 Apr 2007 14:35:38 -0700
On Apr 15, 11:28 am, "NaeiKinDus" <> wrote:

On 15 avr, 01:43, "James Kanze" <> wrote:

On Apr 14, 7:11 pm, "NaeiKinDus" <> wrote:

Hello, i'm trying to program a thread that would be locked (by a
mutex) and that would only be unlocked once that a function
(generating data) is done. The purpose is to generate data, and unlock
the mutex in order to activate the thread once the data is generated.
I have to do it this way, i can only call the thread if the data are

I'm not sure I understand that paragraph. However, there are
one or two things that look more than dubious...

Rephrased: I had to block a thread until the calling process says it
can do what it's meant to do.

You mean you start a thread, just so that it can wait until you
decide it can work. In that case, you need for the calling
thread to lock the mutex before starting the thread, then unlock
it when its OK. The called thread tries acquire the mutex,
blocking until the calling thread releases it. Just don't
forget to have the called thread release the mutex too, when it
is through.

Still, I would definitly consider just putting off starting the
thread until it can do what it's meant to do.

if ((thId = pthread_create(&thread, NULL, OBody::thFileSender, (void
*)params)) != 0)

What are you passing as third pointer? It has to be a global
function, with `extern "C"' linkage. If OBody is a namespace,
this could be OK, but the rest of the code makes me think that
OBody is member function of a class, and there is no way that a
member function of a class can be passed as an argument to a
function, period, and there's also no way that it can have "C"

Nevermind. Just a noob thing a mate did (we're working in group ) -_-'
By the way, why does the function actually works ? It was successfully
called !

I don't know. As I said, to begin with, supposing that
thFileSender is a member function, the code shouldn't even
compile; you can't pass a member function as a parameter,
period, and there's no implicit conversion of member function to
pointer. (I think VC++ did implicitly convert a member function
to a pointer without a &.)

The second thing is that the type of the function is wrong. If
the function is not static, of course, the type is way, way
wrong, and I can't imagine it ever working (but again, I've
heard about some wierd behavior in VC++). But even if the
member function is static, it still won't be `extern "C"'. Now,
Posix doesn't define a C++ binding, and C++ doesn't say anything
about pthread_create, so it's up to the implementation. But in
all the implementations I know (Solaris and Linux),
pthread_create is NOT overloaded, and requires a function with
"C" linkage. The compiler should complain. (Sun CC does,
although it is only a warning, and not an error. G++ doesn't;
this is a bug in g++.)

Anyway... Fixed.

i lock my thread :
void *OBody::thFileSender(void *lpParam)
        cout << "locking MUTX*****" << endl;
        cout << pthread_mutex_lock(params->mx) << endl;
        cout << "unlocked MUTX *****" << endl;
        ... (sending data)

Where do you unlock it? (Your output says "unlocked MUTX", but
it looks to me like you just locked it.) After this operation,
the thread which has executed the pthread_mutex_lock owns the
mutex. No one else can unlock it, for rather obvious reasons.

The "Unlocked" cout was supposed to appear when the thread would have
been released...

Before going any further, you're going to have to read a general
book about threading, at least enough to know the basic
vocabulary. What do you mean by "released"? A thread is
started, it runs, and it exits (terminates). If it hasn't been
detached, someone, somewhere, must join with it. But it's never

James Kanze (Gabi Software) email:
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